| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | State conditions only |
| Difficulty | Moderate -0.8 Part (a) is pure recall of standard Poisson conditions. Parts (b)-(d) are routine calculations using standard Poisson probability formulas and scaling the rate parameter—no problem-solving or novel insight required. This is easier than average for A-level statistics. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | Faults occur randomly, independently |
| and at a uniform average rate | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | Allow constant average rate | Minus 1 mark for |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (b) | P( ≥ 2 faults) = 1 – 0.5249 |
| = 0.4751 | M1 |
| Answer | Marks |
|---|---|
| [2] | Or P(≥ 2 faults) = 1 – P( ≤ 1 fault) |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (c) | Mean = 5 × 1.6 = 8 |
| P(≤ 10 faults) = 0.8159 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | BC | |
| 2 | (d) | Exactly 1 fault in 10 km |
| Answer | Marks |
|---|---|
| P(1 fault) = 0.1304 | M1 |
| Answer | Marks |
|---|---|
| A1 | seen or implied |
| Answer | Marks | Guidance |
|---|---|---|
| In 1 km length, P(0) = 0.7261, P(1) = 0.2324 | M1 | BC (for both) |
| For 10 1 km lengths, P(1) = 10 × 0.72619 × 0.2324 | M1 |
Question 2:
2 | (a) | Faults occur randomly, independently
and at a uniform average rate | E1
E1
[2] | Allow constant average rate | Minus 1 mark for
no context
2 | (b) | P( ≥ 2 faults) = 1 – 0.5249
= 0.4751 | M1
A1
[2] | Or P(≥ 2 faults) = 1 – P( ≤ 1 fault)
BC
2 | (c) | Mean = 5 × 1.6 = 8
P(≤ 10 faults) = 0.8159 | B1
B1
[2] | BC
2 | (d) | Exactly 1 fault in 10 km
So can use Poisson(3.2)
P(1 fault) = 0.1304 | M1
M1
A1 | seen or implied
BC
Alternative solution
In 1 km length, P(0) = 0.7261, P(1) = 0.2324 | M1 | BC (for both)
For 10 1 km lengths, P(1) = 10 × 0.72619 × 0.2324 | M1
= 0.1304
A1
[3]2 A special railway coach detects faults in the railway track before they become dangerous.
\begin{enumerate}[label=(\alph*)]
\item Write down the conditions required for the numbers of faults in the track to be modelled by a Poisson distribution.
You should now assume that these conditions do apply, and that the mean number of faults in a 5 km length of track is 1.6 .
\item Find the probability that there are at least 2 faults in a randomly chosen 5 km length of track.
\item Find the probability that there are at most 10 faults in a randomly chosen 25 km length of track.
\item On a particular day the coach is used to check 10 randomly chosen 1 km lengths of track. Find the probability that exactly 1 fault, in total, is found.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2019 Q2 [9]}}